PA and PB are tangents to a circle with center O, from a point P outside the circle, and A and B are points on the circle. If $$\angle$$APB = $$40^\circ$$, then $$\angle$$OAB is equal to:

As per the given question,

Given that, $$\angle APB=40$$

But we know that tangent always make $$90^\circ$$

Hence, $$\angle AOB+\angle APB=180$$

$$\Rightarrow \angle AOB+40=180$$

$$\Rightarrow \angle AOB=180^\circ-40^\circ=140^\circ$$

$$\Rightarrow \angle OAB=\angle OBA$$

Now, in $$\triangle AOB$$

$$\Rightarrow \angle AOB+\angle OAB+\angle OBA=180$$

$$\Rightarrow 140+2\angle OAB=180$$

$$\Rightarrow 2 \angle OAB=180-140=40$$

$$\Rightarrow \angle OAB=20^\circ$$